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In Triangle Abc, Prove the Following: a 2 ( Cos 2 B − Cos 2 C ) + B 2 ( Cos 2 C − Cos 2 a ) + C 2 ( Cos 2 a − Cos 2 B ) = 0

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Question

In triangle ABC, prove the following: 

\[a^2 \left( \cos^2 B - \cos^2 C \right) + b^2 \left( \cos^2 C - \cos^2 A \right) + c^2 \left( \cos^2 A - \cos^2 B \right) = 0\]

 

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Solution

Let

\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]

Then,
Consider the LHS of the equation

\[a^2 \left( \cos^2 B - \cos^2 C \right) + b^2 \left( \cos^2 C - \cos^2 A \right) + c^2 \left( \cos^2 A - \cos^2 B \right) = 0\]

\[LHS = a^2 \left( \cos^2 B - \cos^2 C \right) + b^2 \left( \cos^2 C - \cos^2 A \right) + c^2 \left( \cos^2 A - \cos^2 B \right)\]
\[ = k^2 \sin^2 A\left( 1 - \sin^2 B - 1 + \sin^2 C \right) + k^2 \sin^2 B\left( 1 - \sin^2 C - 1 + \sin^2 A \right) + k^2 \sin^2 C\left( 1 - \sin^2 A - 1 + \sin^2 B \right) \]
\[ = k^2 \sin^2 A\left( \sin^2 C - \sin^2 B \right) + k^2 \sin^2 B\left( \sin^2 A - \sin^2 C \right) + k^2 \sin^2 C\left( \sin^2 B - \sin^2 A \right)\]
\[ = k^2 \left( \sin^2 A \sin^2 C - \sin^2 A \sin^2 B + \sin^2 A \sin^2 B - \sin^2 B \sin^2 C + \sin^2 C \sin^2 B - \sin^2 C \sin^2 A \right)\]
\[ = k^2 \times 0 = 0 = RHS\]
\[\text{ Hence proved } .\]

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Sine and Cosine Formulae and Their Applications
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